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VladimirAG [237]

3 years ago

12

The temperature outside in the morning was 15 degrees below zero. By lunchtime it was 25 degrees above zero. How many degreees d

id the temperature rise?

1 answer:

denpristay [2]3 years ago

8 0

15 degrees below zero = -15

25 degrees above zero = 25

25 + 15 = 40

It rose 40 degrees.

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Please help me simplify these expressions

Fantom [35]

Answer:

1. -18x¹¹

2. 3n⁷

Step-by-step explanation:

For these problems, there are two things you need to worry about: negative signs and exponents.

1. Let's look at the signs first. There is only one value with a negative sign, meaning that the negative sign will stay.

When multiplying with exponents, you have to add up the exponents. Don't forget the numerical coefficients.

-3x² · 3x · 2x³ · x⁵ = -18x¹¹

2. There are two negative signs in this probem, meaning that they will cancel out. Multiply the rest like we did in the first problem.

3n² · -n² · -n³ = 3n⁷

7 0

3 years ago

This is my question

vovangra [49]

Answer:

week 4 had the greatest weight change

3 0

3 years ago

Organize your data and find the rate of change.

LuckyWell [14K]

A .E is 7 inches for the left arm
E is 8 1/2 inches for the left foot
the inputs look like the length of the arm (ie x) the outputs are the length of the foot
<span>rate of change=<span>change in y change in x</span></span> use the data you gathered or this equation they give you = 0.860 • + 3.302

B. find the difference in y, then the difference in x, then divide (y over x) to get the slope
8-7/7 1/2-6 1/2 = 1/1 =1

C.you plug in any (x,y) point you used to find the slope
Solve for B
y = mx+b y = 1*x + b 7.5 = 1*8 + b b =
7.5 - 8 = -0.5
<span>y = x - 0.5</span>

Hope this helps

8 0

3 years ago

1.The sum of two times a number and 8 is 10. What is the number?<br> A.1<br> B.2<br> C.18<br> D.9

Soloha48 [4]

Answer is Choice A - 1.
1 times 2 is 2, and 2 + 8 is 10.

4 0

3 years ago

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

7 0

3 years ago